Question

Prove \[\frac{sinx}{x} > \frac{2}{\pi} \ \ , \forall x \in (0, \frac{ \pi }{ 2 })\] How to start?

Answer 1

For some reason I'm thinking limits but it's probably because of Sinx / x and has nothing to do with this

Answer 2

I think we need to take limit.. but... we have something to do before taking limit?

Answer 3

Integrate from 0 to pi/2? And something about the area must tell you something? Idk, lol just rambling on.

Answer 4

Well, I don't think we need integration here.

Answer 5

It's because the prof. hasn't taught integration in the lessons yet (even though I have learnt integration before)

Answer 6

Ohh Ok. There are some proof of function being greater than the other.. hmm

Answer 7

1. it's a decreasing function.
2. \[\lim_{x \rightarrow 0} \frac{sinx}{x}=1\]
3. \[\lim_{x \rightarrow \frac{\pi}{2}} \frac{sin\frac{\pi }{2}}{\frac{\pi}{2}}=\frac{2}{\pi}\]

Uh-oh! I'm not on the right track :|

Answer 8

Why can't you just show that it is uniformly decreasing over the interval and calculate the least value? (Once you have shown that the limit is 1 at start)

Answer 9

Well, that's the problem.. How are you going to show it? Finding its derivative?

Answer 10

you cannot just say that the roots are at n*pi?

Answer 11

How so?

Answer 12

You want a proof of that as well...

Answer 13

Just a minute
\[\frac{d}{dx}\frac{sinx}{x} = \frac{xcosx-sinx}{x^2}\]If I put it =0, I'll get
xcosx - sinx = 0
That's where I stuck at.

And I haven't learnt the general solution (hmm, for that n*pi) yet, so I would like to know how you get that.

Answer 14

I mean x= Tan x is a pain.....

Answer 15

Yes!

Answer 16

You just have do a graph y=x and Tan x and look or use a CAS

Answer 17

Graphing... is not that good...

Answer 18

If graphing works, why not just graph the function sinx / x to show it's a decreasing function?

Answer 19

Well, quite...:-)

Answer 20

What are you doing, Fourier?

Answer 21

No.. limit and derivative.

Answer 22

That's a bit sneaky, hitting you with sinc function...

Answer 23

Let's see , all you have to do is show that the first positive root is at pi....

Answer 24

Sin pi/pi = 0

Answer 25

And anywhere from 0 to pi is not 0

Answer 26

That's good enough to show the first root is at pi, right?

Answer 27

Oh... so you're solving sinx/x=0? If so, yes!

Answer 28

So now you are done, I think (assuming usual smoothness principles).

Answer 29

I don't see the problem by saying that the limit at 0 is 1>pi/2, and that sinx/x is uniformly decreasing over (0,pi/2), and the limit approaching pi/2=2/pi

that shows that sinx/x>2/pi in that region

Answer 30

So, step 1, 2, 3 are correct. But how to show step 1 is the point. I'm not quite sure if I can get estudier's idea..

Answer 31

how to show it's decreasing? find critical points\[f'=\frac{x\cos x-\sin x}{x^2}=0\implies x\cot x=0\]so the critical points are\[x=\{0,\frac\pi2\}\]hence you just need to test a point in that region, say \(x=\pi/4\), and that tells you whether the function is increasing or decreasing on that interval

Answer 32

Wait.. how.. why...
xcosx - sinx = 0
xcotx -1 = 0
xcotx = 1
Am I doing anything wrong here?

Answer 33

oh shnap I am out to lunch, but I still say we find that critical point and just test the interval
let me think how....

Answer 34

Don't think while you eat!

Answer 35

I know I have seen a proof that tanx>x in (0,pi/2) ... that would work if I could remember it

Answer 36

This is complicated...

Answer 37

I got it :)
so are you with me up to the idea that tanx>x in (0,pi/2) would prove that this is decreasing?

Answer 38

Yes!

Answer 39

okay, we can prove this using the mean value theorem...

Answer 40

the mean value theorem states that f(x)=f(a)+f'(c)(x-a) for some a<c<x
our f(x)=tanx, and our interval is (0,pi/2) so a=0

Answer 41

getting what we need from the formula
f(0)=0, f'(c)=sec^2(c) where 0<c<x<pi/2

Answer 42

what do you get then by using this in the formula?

Answer 43

f(x)=f(a)+f'(c)(x-a)
tanx = 0 + sec^2 (c) (x)
tanx = [sec^2 (c)] x
Doesn't look good.

Answer 44

it is though, because what do we know about sec(c) for 0<c<pi/2 ???

Answer 45

0<c<pi/2
1<sec (c)< undefined (+infinity?!)

Answer 46

yeah, but the point is that sec(c)>1 for that interval

Answer 47

so
tanx=sec^2(c)x 0<c<pi/2 implies that...?

Answer 48

1< sec^2(c)
x < sec^2 (c) x
x< tanx
Wow!!

Answer 49

tadah! :)
so that completes your proof

Answer 50

Uh-huh! Why did you think of mean value theorem for this question????

Answer 51

It's the only way I know to prove tanx>x in (0,pi/2)
there may be other ways

Answer 52

Given the interval is \( (0, {\pi \over 2}) \)
to find the maximum or minimum, we have
\[ x = \tan x\]
the first part is a line y=x passing through origin while the latter one is y=tan(x)

we can see that tan(x) > x since tan(x) > slope is sec^2(x) > slope of (x) = 1

so next time tan(x) meets x in the interval pi/2, pi

Hence we do not have any critical point in the interval (0, pi/2)